A) 16 eV
B) 3.45 keV
C) 3.47 keV
D) 3.43 keV
Correct Answer: C
Solution :
\[{{E}_{K}}-{{E}_{M}}=\frac{1242}{0.36}=3450eV\] (Energy corresponding to\[{{K}_{\beta }}X-ray\]) \[{{E}_{M}}-0=16eV\] (lionization energy) from M shell. The energy needed to knock out an \[{{e}^{-1}}\] from K shell is \[{{E}_{K}}=3450+{{E}_{M}}\] \[{{E}_{K}}=3466eV\] \[=3.466keV\approx 3.47keV\]You need to login to perform this action.
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